Question

Show that the given differential equation is homogenous and solve them
$\frac{dy}{dx}=\frac{x+y}{x-y}$

Answer 1

$\frac{dy}{dx}=\frac{x+y}{x-y}$

Let $x=ax$ and $y=ay$

$\frac{dy}{dx}=\frac{ax+ay}{ax-ay}=\frac{x+y}{x-y}$

$\therefore$ Homogeneous differential $e{q}^n$

Now let $y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$

$v+x\frac{dv}{dx}=\frac{x+vx}{x-vx}=\frac{1+v}{1-v}$

$\Rightarrow x\frac{dv}{dx}=\frac{1+v}{1-v}-v=\frac{1+v-v(1-v)}{1-v}$

$\Rightarrow x\frac{dv}{dx}=\frac{1+v^2}{1-v}\Rightarrow \frac{1-v}{1+v^2}dv=\frac{dx}{x}$

$\Rightarrow (\frac{1}{1+v^2}-\frac{v}{1+v^2})dv=\frac{dx}{x}$

integrating both sides

$ta{n}^{-1}v-\frac{1}{2}log(1+v^2)=logx+c$

$\Rightarrow ta{n}^{-1}\left(\frac{y}{x}\right)+log\left(\frac{1}{\sqrt{1+v^2}}\right)-logx=C$

$\Rightarrow ta{n}^{-1}\left(\frac{y}{x}\right)+log\left(\frac{x^2}{\sqrt{x^2+y^2}}\frac{1}{x}\right)=C.$

$\Rightarrow ta{n}^{-1}\left(\frac{y}{x}\right)+log\left(\frac{x}{\sqrt{x^2+y^2}}\right)=C.$